Timeline for Polynomial satisfying a functional equation [closed]
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 9, 2020 at 15:17 | history | closed |
user44191 Alex M. ARG Ben McKay LeechLattice |
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| May 9, 2020 at 10:55 | comment | added | ARG | You can solve the problem by looking at the smallest (or largest coefficient first) and then working your way up (or down). For example, by setting $x=0$ you get $p(0) = 1/2q(0)$. So you can try to work the coefficients of $p$ one by one. For an abstract argument, consider $a_i$ to be the coefficients of $p$ (you have $n+1$ of them). Then $F(x) = q(x)p(x)$ has degree $2n+1$. When looking at $F(x) + F(-x)$ all monomials of odd degree cancel. There remains $n+1$ even degree monomials. So $n+1$ linear constraints and $n+1$ variables. There is a solution (use distinct roots for uniqueness). | |
| May 6, 2020 at 23:56 | review | Close votes | |||
| May 9, 2020 at 15:17 | |||||
| May 6, 2020 at 21:24 | history | became hot network question | |||
| May 6, 2020 at 13:51 | answer | added | abx | timeline score: 7 | |
| May 6, 2020 at 13:27 | history | edited | Martin Sleziak |
added the (functional-equations) tag
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| May 6, 2020 at 13:24 | history | asked | user157299 | CC BY-SA 4.0 |